Question: Let $f(x)=2x^3-9x^2+12x$ Where does $f$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=1$ (Choice C) C $x=2$ (Choice D) D $f$ has no critical points.
A critical point of $f$ is a point in the domain of $f$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $f$, let's find its derivative. $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[ 2x^3-9x^2+12x \right] \\\\ &=6x^2-18x+12 \\\\ &=6(x^2-3x+2) \\\\ &=6(x-1)(x-2) \end{aligned}$ Now let's look for $x$ -values where $f'$ is zero or undefined. $6(x-1)(x-2)=0$ at $x=1$ and at $x=2$. $6(x-1)(x-2)$ is never undefined, so $f'$ is never undefined. In conclusion, these are the $x$ -values where $f$ has critical points: $x=1$ $x=2$